Q74 / 160 · JIPMER 2009 · Chemistry

$9.2 \mathrm{~g} \mathrm{~N}_2 \mathrm{O}_4$ is heated in a 1 L vessel till equilibrium state is established $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$ In equilibrium state $50 \% \mathrm{~N}_2 \mathrm{O}_4$ was dissociated, equilibrium constant will be (mol. wt. of $\mathrm{N}_2 \mathrm{O}_4=92$ )

  • A
    0.1
  • B
    0.4
  • C
    0.3
  • Correct answer
    D
    0.2

Solution

$\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})$ Molar concentration of $\left[\mathrm{N}_2 \mathrm{O}_4\right]=\frac{9.2}{92}=0.1 \mathrm{~mol} / \mathrm{L}$ In equilibrium state when it 50\% dissociates, $\begin{aligned} {\left[\mathrm{N}_2 \mathrm{O}_4\right] } & =0.05 \mathrm{M} \\ {\left[\mathrm{NO}_2\right] } & =0.1 \mathrm{M} \\ K_c & =\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2 \mathrm{O}_4\right]} \\ & =\frac{0.1 \times 0.1}{0.05}=0.2 \end{aligned}$

Hence, the correct answer is option: 'D'.

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